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# A solid floats in a liquid at $20^\circ {\text{C}}$ with $75\%$ of its volume immersed. When the liquid is heated to $100^\circ {\text{C}}$, the same liquid floats with $80\%$ of its volume immersed in the liquid. Calculate the coefficient of expansion of the liquid (Assume the volume of the solid to be constant)1) $8.33 \times {10^{ - 4}}\;/^\circ {\text{C}}$ 2) $83.3 \times {10^{ - 4}}\;/^\circ {\text{C}}$3) $833 \times {10^{ - 4}}\;/^\circ {\text{C}}$ 4) $0.833 \times {10^{ - 4}}\;/^\circ {\text{C}}$

Last updated date: 15th Aug 2024
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Hint: The above problem can be solved by using the concept of thermal expansion. The thermal expansion is the variation in the size of the object due to variation in temperature of the object. The thermal expansion may be linear expansion, surficial expansion and volumetric expansion. The expansion of the object depends on the temperature and coefficient of expansion.

Given: The initial temperature of the liquid is ${T_1} = 20^\circ {\text{C}}$, the final temperature of the liquid is ${T_2} = 100^\circ {\text{C}}$, the immersed volume of the object at initial temperature is ${V_1} = 0.75V$, the immersed volume of the object at final temperature is ${V_2} = 0.8V$.
$\gamma = \dfrac{{{V_2} - {V_1}}}{{{V_1}\left( {{T_2} - {T_1}} \right)}}$
Substitute $0.8V$for ${V_2}$ , $0.75V$for ${V_1}$ , $20^\circ {\text{C}}$ for ${T_1}$ and $100^\circ {\text{C}}$ for ${T_2}$ in the above expression to find the coefficient of expansion of the liquid.
$\gamma = \dfrac{{0.8V - 0.75V}}{{\left( {0.75V} \right)\left( {100^\circ {\text{C}} - 20^\circ {\text{C}}} \right)}}$
$\gamma = 8.33 \times {10^{ - 4}}\;/^\circ {\text{C}}$
Thus, the coefficient of expansion of the liquid is $8.33 \times {10^{ - 4}}\;/^\circ {\text{C}}$ and option (a) is the correct answer.